Line ℓ has equation y=x+3. Find the distance between ℓ and the point T( - 1,5)

Answer 1

y=x+3

(-1,5)

the distance between the line and the point above is found by finding a line perpendicular to the given line and through the point. then find the intersection of the two lines. then calculate the distance from that intersection point to (-1,5)

y=x + 3 has slope = 1, the coefficient of the x term, when the equation is solved for x a perpendicular line has the negative inverse as the slope or -1

y=-x + b, solve for b by plugging in the point

5=-(-1) + b

b = 4

y=-x +4 is the perpendicular line

it intersects y=x+3

when -x+4 = x+3

2x = 1

x= 1/2

y= 3.5= 7/2

calculate the distance from (1/2, 7/2) to (-1, 5)

use the distance formula

d^2 = (5-7/2)^2 + (-1-1/2)^2 = 9/4 + 9/4 = 18/4 = 9/2

d = sqr(9/2) = 3/sqr2 = about 3/1.41 =about 2.12